Question

4. A line segment gets divided by the point
(2k - 3/k+1, k +3/k+1)
in the ratio k: 1. If point
of division lies on the x-axis then its coordinates are a.(9/2,0) b.(-9/2,0) c.(0,9/5) d.(0,-9/5)​

Answer 1

(9/2 , 0) is the point  if A line segment gets divided by the point (2k - 3/k+1, k +3/k+1) in the ratio k: 1. If point of division lies on the x-axis

Step-by-step explanation:

point (2k - 3/k+1, k +3/k+1) divides the line segment

Point of Division lies on x axis

=> y = 0

=> (k + 3)/(k + 1) = 0

=> k + 3 = 0

=> k = - 3

Point  =  (2k - 3)/(k + 1)  =  ( -6 - 3)/(-3 + 1)

= -9/(-2)

= 9/2

(9/2 , 0) is the point

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Answer 2

Therefore the coordinate of the divisor point is $=(0,\frac{9}{5})$

Step-by-step explanation:

Given , A line segment get divided by a point $(\frac{2k-3}{k+1} ,\frac{k+3}{k+1} )$ in the ratio k:1 and point of divisor lies on x-axis .

Since the point of divisor lies on the x-axis , then the x- coordinate of the point will be zero.

$\therefore \frac{2k-3}{k+1} =0$

$\Leftrightarrow 2k-3 = 0$

$\Leftrightarrow k= \frac{3}{2}$

Putting the value of k in the given point we get

$(0,\frac{\frac{3}{2} +3}{\frac{3}{2} +1} )$   $=(0,\frac{\frac{9}{2} }{\frac{5}{2} })$   $=(0,\frac{9}{5})$

Therefore the coordinate of the divisor point is $=(0,\frac{9}{5})$